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Find the derivative of the function.

  1. Feb 22, 2005 #1
    Can you work this out step by step so I can see how to do it. Thank you
    h(t) = (t^4 - 1)^3(t^3+1)^4
     
    Last edited: Feb 22, 2005
  2. jcsd
  3. Feb 22, 2005 #2

    saltydog

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    It's a bit ambiguous. Do you mean:

    [tex](t^4-1)^3 (t^3+1)^4[/tex]

    I think so. In that case, need to use the Chain Rule. If it confussing to you, try a simpler problem first (I do that too). For example, try this one:

    [tex](t^2-1)t^3[/tex]

    Wouldn't that just be:

    [tex](t^2-1)3t^2+t^3(2t)[/tex]
     
  4. Feb 23, 2005 #3
    Well you're going to use the product rule, so first use the chain rule to get the derivative of each factor:

    [tex] \left[ (t^4 - 1)^3 \right]^{\prime} = 3(t^4 - 1)^2 4t^3 = 12t^3(t^4-1)^2[/tex]

    [tex]\left[ (t^3 + 1)^4 \right]^{\prime} = 4(t^3 + 1) 3t^2 = 12t^2(t^3 + 1)^3 [/tex]

    Then apply the product rule:

    [tex]12t^3(t^4-1)^2 (t^3 + 1)^4 + 12t^2(t^3 + 1)^3 (t^4 - 1)^3 [/tex]
     
  5. Feb 23, 2005 #4

    Tom Mattson

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    We don't do that here. At Physics Forums you have to show your work to receive homework help.
     
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